It's not homework; I'm reading that book recreationally and these are only problems in these chapters I'm stuck on. I can wait longer for an answer. Otherwise I agree that homework should be tagged.
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sdcvvcOct 11 '09 at 15:04

4

Perhaps we should have a tag instead for "exercise"s or "book-problem"s.
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Scott Morrison♦Oct 11 '09 at 18:38

Yeah, but if a tag is too narrow it is useless. I think that "exercise in a graduate level text" is about the right level of specificity, and I am fine with calling it "exercise".
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David SpeyerDec 1 '09 at 17:01

5 Answers
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For any real number x, let f(x) be the real obtained by interleaving the binary digits of x with the binary digits of pi, a fixed irrational number. This is clearly order-preserving, and f(x) is irrational since the digits will not repeat. QED

Minor nit-pick: One has to choose a 2-adic expansion of every $x$ to do this, for example the (uniquely determined) one that has infinitely many ones. It matters because the digit sequences 100000... and 0111111... will be mapped to different irrationals even though they represent the same rational number.
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Johannes HahnJan 25 at 17:07

Yes; so just use your favorite representation (say, the one without tailing $1$s), and there is no problem.
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Joel David HamkinsJan 25 at 17:24

(3) This is only a minute variant on Ilya's answer, but I think it looks a bit more perspicuous in continued fractions.
In fact, there's a negative continued fraction expansion that Richard Guy told me about which have a nicer ordering than standard regular cfs. Every real is uniquely representable as an infinite continued fraction
$x=a_{0}-1/\left(a_{1}-1/\left(a_{2}-1/\left(\ldots\right)\right)\right)$ with all $a_{n}$ in $\mathbb {Z}$ and $a_{n}\ge 2$ for $i\ge 1$. $x$ is rational iff the all $a_{n}$ are $2$ from some point on. And the negative continued fractions have the lexicographic order: $x\gt x^{\prime}$ iff $a_{i}\gt a^{\prime}_{i}$ for the first $i$ for which $a_{i}$ and $a^{\prime}_{i}$ differ.

So just map $x$ to $\left(a_{0}+1\right)-1/\left(\left(a_{1}+1\right)-1/\left(\left(a_{2}+1\right)-1/\left(\ldots\right)\right)\right)$.

1) modify Cantor function so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.

2) consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.

3) if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long Ehrenfeucht–Fraïssé game). But they are not. I think. (well, it's been shown that such embedding is actually possible, so my assumption was incorrect)

1) Take any nondecreasing continuous function from the reals to the reals that is constant in neighborhoods of rationals, and restrict it to irrationals. This can be constructed as a uniform limit by starting with f(0)(x) = x, enumerating the rationals as r(i) and for each i, setting f(i+1)(x) to be a sum of f(i)(x) and some piecewise linear function supported in a neighborhood of radius 2^(-i) around r(i).

The finite interval assertion in the last sentence is clearly false, since you can take an infinite subsequence of rationals with nested neighborhoods. I don't see a way to rescue the argument (but it's more or less moot at this point).
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S. Carnahan♦Oct 11 '09 at 19:05

(3) Here's how to construct an example. We can assume the segment in question was from 0 to 1 non-inclusively. Also, I will write source numbers in binary and target numbers in base 4.

Consider first the number 1/2, in binary 0.1000... Let's map it to 0.1[0102010...] — it doesn't matter what's inside [...] as long as it's irrational. Now we decide to map all numbers of the form 0.0... to 0.0... and 0.1... to 0.2... . Clearly, so far we didn't break anything.

Now, let's take a different rational number, e.g. 1/4 = 0.010... Similarly, we decide to map it to one of irrationals of the form 0.01[10011010...] and the segment [1/4, 1/2] is ready to go to 0.02...

Select the next rational number, e.g. 1/3 = 0.0101(01). It's breaking in half the segment destined to go to 0.02... No problem, again we select some irrational 0.021[010012...] for 1/3 and move left and right subsegments to 0.020... and 0.022...

Now, so far I was using xxx0, xxx1 and xxx2. But let's sometimes move segments to xxx1, xxx2 and xxx3. Let's do it whenever I'm on a level which is a square of natural number.

We're still increasing, yahoo!

Repeat this process for all rationals ordered by denominator. For any rational we have selected an irrational number by definition. For any irrational, it's the limit of segments broken down into parts. Each breakdown reveals exactly one digit of the result — so we reconstruct it digit-by-digit in ternary. It has infinite number of digits. Moreover, these digits never become periodical thanks to the fact that each $n^2$-digit was shifted by 1. So the result is irrational too.

Since every two irrationals are separated by rational, this function is always increasing. Qued erat construirum.